i’m currently figuring out how to do a quite simple Task, but i cant get my head around it. I’ve got the Intel XDK installed, added a new Project and chose the Ionic Framework for it.
Then I have setup 3 simple Tabs with the APP Designer Function. Website | Facebook | Twitter
(as the drag and drop is quite nice) Now I have got those 3 different Tabs and an empty window under it, where I want the external Links to be opened in. I have the Setup so far working, that when i click those buttons it opens those links in a new window.
<a class="tab-item widget uib_w_2" data-uib="ionic/tab_item" data-ver="0" onClick="window.open('http://www.example.com','_self','location=yes');return false;">
However I would like to have it so that when I Click on the Tab it should open the link in the same window beneath the Tabs, so that the Tabs stay and are further clickable for switching. Currently im only getting forwarded to the specified Page and the Tabs just simply hide, which is not what i want.
Do I have to change something small for that to work or am I understanding everything wrong?
I have the InAppBrowser plugin installed aswell and have used all arguments from _self over _blank. Doesnt make a difference for what im needing. Could someone please point me to the right Direction or give me the Code Snippet which does achieve this Function? I hope this hasnt been asked a thousand times before, if so please pardon me as English isnt my native Language.